In the adiabatic compression, the decrease in volume is associated with

One mole of an ideal gas expands adiabatically at constant pressure such that its temperature $T \propto \frac{1}{\sqrt{V}}$. The value of $\gamma$ for the gas is $(\gamma = \frac{C_p}{C_v}, V = \text{Volume of the gas})$

$A$ polyatomic gas is compressed to $\left(\frac{1}{8}\right)^{\text{th}}$ of its original volume adiabatically. If its initial pressure is $P_0$,what will be its new pressure (in $P_0$)? (Given: $\gamma = \frac{C_p}{C_v} = \frac{4}{3}$)

$A$ monoatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume is $TV^{x} = \text{constant}$, then $x$ is

$A$ monoatomic ideal gas, initially at temperature $T_1$, is enclosed in a cylinder fitted with a massless, frictionless piston. By releasing the piston suddenly, the gas is allowed to expand adiabatically to a temperature $T_2$. If $L_1$ and $L_2$ are the lengths of the gas columns before and after expansion respectively, then $(T_2 / T_1)$ is given by

$A$ monoatomic gas is stored in a thermally insulated container and the gas is suddenly compressed to $(\frac{1}{8})^{\text{th}}$ of its initial volume. The ratio of final pressure to initial pressure is (in $:1$)